Here's how I figured it.
x^2-3x is greater than or equal to 10.
I subtracted 10 from both sides. This left me with x^2-3x-10 is greater than or equal to 0. Next, I factored and came up with (x-5)(x+2) is greater than or equal to 0. Logically, it would follow that x=5 and x= -2 would it not? And then in symbolic form my answer would read x is less than or equal to -2 OR x is greater than or equal to 5.

The part " x=5 and x= -2" is only part of the solution, namely the equality part.
You still have to solve the inequality part.

There are two ways to look at this.
First, the easy way witout any calculatoins. For a parabola with the leading coefficient (1)x² >0, we know that the curve is concave upwards, so the values of x between these two zeroes must be less than zero. The answer should therefore be:
(-∞,-2]∪[5,∞), or
x≤-2 or x≥5.

The second way is to assign values of x close to but not equal to the zeroes.
f(-10) = 120
f(-2.1) = 0.71
f(-1.9) = -0.69
f(0) = -10
f(4.9) = -0.69
f(5.1) = 0.71
f(10) = 60
It is clear then that for f(x) = x²-3x-10 ≥ 0,
x ≤ -2 or x ≥ 5

Please don't start a new post for a question if you have a second part to the same problem.
Also give yourself a "First Name", it is much easier to keep track of replies.

Now to your question,
you had
(x-5)(x+2) < 0
I showed you a simple way to proceed from there, but apparently you did not understand it.

so here is the "formal" way
We are multiplying two factors and the result is negative ( < 0)
so one must be positive , the other negative but we don't know which one, so

1. x-5 > 0 AND x+2< 0
x>0 AND x < -2

OR

2. x-5<0 AND x+2>0
x<5 AND x>-2

clearly the first case is not possible,
Can you think of a number which is greater than 2 AND less than -2 ? There is none

Since the two cases are separated by OR, the second case is true.
now recall that the symbold was ≤ and ≥ so we have to include the -2 and 5
and my solution would be all values between -2 and 5 inclusive or
-2 ≤ x ≤ 5, as I stated earlier.